substituting the first equation into the second: $$3xy+(x-y)^2(x+y) = (x-y)^2(x+y)^2$$so substituting $a=x+y,b=x-y$ and knowing that $4xy=a^2-b^2$: $$\frac{3a^2-3b^2}{4} + ab^2 = a^2b^2$$$$3a^2-3b^2+4ab^2=4a^2b^2$$$$3a^2=b^2(4a^2-4a+3)$$$$b^2= \frac{3a^2}{4a^2-4a+3}$$giving two cases: $a = 0 \rightarrow b = 0 \rightarrow (x,y) = (0,0)$ Or: $$3a^2 \ge 4a^2-4a+3$$$$0 \ge (a-1)(a-3)$$since a is an integer this means $a=\{1,2,3\}$ checking each of these gives the solutions $(x,y) = (1,0), (2,1)$. So all solutions are: $(x,y) = (0,0), (1,0), (2,1)$ edit: just spotted the solution (0,-1) so i will look for the problem with my proof.

Continuing the above from $a \in \{1, 2, 3\}$, we have $(a, b)=(1, \pm 1), (3, \pm 1)$, and the $(0,0)$ from earlier. This gives the solutions: $$(x, y, z)=(2, 1, 3), (1, 2, -3), (1, 0, 1), (0, 1, -1), (0, 0, 0)$$

furioushatter wrote: Continuing the above from $a \in \{1, 2, 3\}$, we have $(a, b)=(1, \pm 1), (3, \pm 1)$, and the $(0,0)$ from earlier. This gives the solutions: $$(x, y, z)=(2, 1, 3), (1, 2, -3), (1, 0, 1), (0, 1, -1), (0, 0, 0)$$ Yh sorry that was a silly slip.

solution

Substitute the first equation into the second equation to get that $$3xy+(x-y)(x^2-y^2)=(x^2-y^2)^2.$$Let $a = x + y$ and let $b = x-y$. Then, we know that $a^2 = x^2+y^2 +2xy$ and $b^2 = x^2+y^2-2xy$, so $a^2-b^2 = 4xy$, which means that $3xy = \frac{3(a^2-b^2)}{4}$. We can re-write the new equation as: $$\frac{3(a^2-b^2)}{4}+b^2a=a^2b^2 \Longleftrightarrow b^2 = \frac{3a^2}{4a^2-4a+3}$$If $a = 0$, then $b=0$ (which is the same thing as $x=y=0$, so for now let’s assume that $a \ne 0$. Then, we want $$3a^2 \ge 4a^2-4a+3 \Longleftrightarrow a^2-4a+3 \le 0 \Longleftrightarrow (a-1)(a-3) \le 0.$$This means that we only need to test $a = 1,2,3$. if $a = 1$, then $b = \pm 1$. This produces two solutions $(x,y)$: $(1,0)$ and $(0,1)$. If $a =2 $, then $b$ isn’t an integer. If $a = 3$, then $b= \pm 1$, which produces two solutions $(x,y)$: $(2,1)$ and $(1,2)$. Therefore, the only solutions are $(x, y, z)=\boxed{(2, 1, 3), (1, 2, -3), (1, 0, 1), (0, 1, -1), (0, 0, 0)}$.